Дыхта В.А. Динамические системы в экономике. Введение в анализ одномерных моделей
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y(t) =
1
2
+ (C −
1
2
) 2
t
, C = y
0
,
y(t) = C
1
(
√
2)
t
+ C
2
(−
√
2)
t
− 6, t = 0, 1 . . . ,
y
t
y(t)
y
n
n = 0, 1, . . .
y
0
= f(t, y), (1.6)
f(t, y)
h > 0
y
0
∆y
h
=
y(t + h) − y(t)
h
(1.7)
∆y/∆t = f(t, y) ∆t = h
y(t + h) = y(t) + hf (t, y(t)). (1.8)
y(t
0
) y(t
0
+ h)
y(t
0
+ 2h) h → 0
y
00
= f(t, y, y
0
). (1.9)
y
0
y
00
∆
2
y
h
2
=
y(t + 2h) − 2y(t + h) + y(t)
h
2
, (1.10)
∆
2
y
(∆t)
2
= f
µ
t, y,
∆y
∆t
¶
,
y(t + 2h) − 2y(t + h) + y(t)
h
2
= f
µ
t, y,
y(t + h) − y(t)
h
¶
(1.11)
y h
y h
y(t
0
+ nh)
h → 0
n
n
0, 1, 1, 2, 3, 5, 8, 13, . . .
F
n+2
= F
n+1
+ F
n
, (1.12)
t
0
t
0
+h t
0
+nh t
0
+Nh
t
y
0
y
0
= ay + b, (1.13)
y(t + 1) = ay(t) + b. (1.14)
a b
b = 0
y
0
/y = a
y(t)
a
y a
y(0) = y
0
a
y(t) = y
0
e
at
a > 0
a > 0
y
0
= ay
t
y
0
t
y
0
y
0
= ay a > 0
a < 0
b = 0
y(t+1)/y(t) = a
a = a(t)
S
0
r
S(t) t
S(t)
S(t + 1) = (1 + r)S(t), t = 0, 1, 2, . . . (1.15)
S(0) = S
0
t
S(t) = S
0
(1 + r)
t
.
r%
S(t + 1) = (1 +
r
2
)S(t),
t =
0, 1/2, 1, 3/2, . . . 1/2
S(t) = S
0
³
1 +
r
2
´
2t
,
t
m
1/m
S(t + 1) =
³
1 +
r
m
´
S(t), (1.16)
t = 0, 1/m, 2/m, . . . 1/m
t
m
S(t) = S
0
³
1 +
r
m
´
mt
, (1.17)
m → ∞
m
S(t) = lim
m→∞
S
0
³
1 +
r
m
´
mt
= S
0
e
rt
(1.18)
h = 1/m
h → 0 m → ∞
S(t + h) − S(t)
h
= rS(t).
m → ∞ h → 0
dS/dt
h
dS(t)
dt
= rS(t) S
0
= rS. (1.19)
S(t)
S(0) = S
0
r = 8% r = 0, 08
r = 0, 08
S(t)/S
0
S(t)/S
0
m = 4 m = 366
S
0
= 1000
r
c(t)
t
S(t + 1) = (1 + r)S(t) + c(t). (1.20)
c(t)
c(t)
c(t)
c(t) > 0
c(t) < 0 c(t) = c =
S
0
(t) = rS(t) + c(t) (1.21)
S(0) = S
0
c(t)
c(t) = c =
S(t) = S
0
e
rt
+
c
r
¡
e
rt
− 1
¢
.
w(t)
c(t)
S(t)
S
0
(t)
w(t) + rS(t) = c(t) + S
0
(t) .
S
0
S
0
(t) = rS(t) + w(t) − c(t). (1.22)
w(t) c(t)
•
M
d
P
= e
−απ
, (1.23)
P π
α > 0
•
π
0
= β
µ
P
0
P
− π
¶
, (1.24)
β > 0 P
0
/P
π
0
> 0
m
d
= ln M
d
, p = ln P.
m
d
− p = −απ,
π
0
= β(p
0
− π), (1.25)
M
d
= m
d
= p
t
p
0
= απ
0
π
0
= γπ, γ =
β
αβ − 1
.
π(0) = π
0
π(t) = π
0
e
γt
γ < 0
αβ < 1 π(t) → 0 t → ∞
y
0
= ay − by
2
, (1.26)
y(t + 1) = ay(t) − by
2
(t), (1.27)
a, b > 0
y
0
y
= a − by,
y(t + 1)
y(t)
= a − by(t),
y
y
ay y
y
y
y
0
y
= ϕ(y), y
0
= ϕ(y)y,
ϕ(y) ≈ a > 0 y
ϕ(y) < 0 y
r
r
³
1 −
S(t)
S
max
´
,